Let the sum of the first n terms of infinite arithmetic sequence {an} be SN. If the inequality an ^ 2 + Sn ^ 2 / N ^ 2 > = (T / 5) A1 ^ 2 holds for any positive integer n, then the maximum value of real number T is () A.1 B.2 C.3 D.5 A little thought Ideas

Let the sum of the first n terms of infinite arithmetic sequence {an} be SN. If the inequality an ^ 2 + Sn ^ 2 / N ^ 2 > = (T / 5) A1 ^ 2 holds for any positive integer n, then the maximum value of real number T is () A.1 B.2 C.3 D.5 A little thought Ideas

In reply I'm in a hurry, so I can only solve it in this way for the time being. I hope I can understand
In the formula, there are four variables, three of which are related, and the absolute value relationship between N and the other three is not strong
Consider killing it first
From the summation formula of arithmetic sequence, Sn = [(a1 + an) / 2] (n) (emphasize that there is a coefficient n), we know that the square of N can be reduced
After that, we use the general term formula of arithmetic sequence to expand it
The left side of the original formula = 2A1 ^ 2 + (5 / 4) (n-1) ^ 2 D ^ 2 + 3A1 (n-1) d
There is a 2A1 ^ 2 in the front, so you can look at the back
The minimum value depends on how small it is
Take D as independent variable, the original formula is quadratic function, opening up
From (4ac-b ^ 2) / 4A (the most value formula of quadratic function), we can be surprised to find that this value is (- 9 / 5) a ^ 2
Add in the front two and you'll have to
The original minimum is (1 / 5) a ^ 2
Then the value of T is a
The problem is not difficult It takes less time than typing
It's better to try to solve this kind of problem first
above

The known sequence {log2 ^ an} (n is a positive integer) is an arithmetic sequence, A1 = 2, A3 = 8
Given that the sequence {log2 ^ an} (n is a positive integer) is an arithmetic sequence, A1 = 2, A3 = 8, (1) find the general term formula of the sequence {an}, (2) find the sum of the first n terms of the sequence {1 / an} as Sn, find the sum of the first n terms of the sequence {NSN} and TN

(1) Let the tolerance of sequence {log2 (an)} be dlog2 (A3) - log2 (A1) = 2D = log2 (8) - log2 (2) = 3-1 = 2D = 1log2 (an) = log2 (A1) + (n-1) d = log2 (2) + (n-1) = 1 + n-1 = Nan = 2 & # 8319; when n = 1, A1 = 2; when n = 3, A3 = 2 & # 179; = 8, the general formula of sequence {an} is an = 2 & # 8319; (...)

In known sequence {an}, A1 = 1, an = (an-1) (3 ^ n-1) (n > = 2 and N belongs to positive integer)
Finding the general term formula of sequence {an}

In ∵ sequence {a [n]}, a [n] = a [n-1] 3 ^ (n-1) (n ≥ 2, and N ∈ positive integer) ∵ a [n] / a [n-1] = 3 ^ (n-1) a [n-1] / a [n-2] = 3 ^ (n-2)... A [3] / a [2] = 3 ^ 2A [2] / a [1] = 3 ^ 1. Multiply the above expressions to get: a [n] / a [1] = 3 ^ [1 + 2 +... + (n-1)] ∵ a [1] = 1A [n] = 3 ^ [n (n-1) / 2]